# Does Nakayama Lemma imply Cayley-Hamilton Theorem?

Consider the Cayley-Hamilton Theorem in the following form:

CH: Let $A$ be a commutative ring, $\mathfrak{a}$ an ideal of $A$, $M$ a finitely generated $A$-module, $\phi$ an $A$-module endomorphism of $M$ such that $\phi(M)\subseteq\mathfrak{a}M$. Then there are coefficients $a_i\in\mathfrak{a}$ such that $\phi^n+a_1\phi^{n-1}+\dots+a_n=0$.

This theorem can be proved by using elementary linear algebra in the context of rings. As a corollary, we find the following two versions of Nakayama's Lemma:

NAK1: Let $A$ be a commutative ring, $M$ a finitely generated $A$-module and $\mathfrak{a}\subseteq A$ an Ideal such that $\mathfrak{a}M=M$. Then there is an $x=1\mod\mathfrak{a}$ such that $xM=0$.

Proof. One just sets $\phi=\operatorname{id}$ and plugs in $x=1+a_1+\dots+a_n$.

It follows:

NAK2: Let $M$ be a finitely generated $A$-module, $\mathfrak{a}$ an ideal contained in the Jacobson radical of $A$. Then $\mathfrak{a}M=M$ implies $M=0$.

Proof. Indeed, $xM=0$ for an element $x\in 1+\mathfrak{a}\subseteq 1+J(A)$, which is a unit, hence $M=0$.

However, one can prove Nakayama's Lemma avoiding linear algebra:

Alternative proof of NAK2: Let $u_1,\dots,u_n$ be a generating system of $M$. $u_n\in M=\mathfrak{a}M$, so $u_n=a_1u_1+\dots+a_nu_n$. Subtracting, $(1-a_n)u_n=a_1u_1+\dots+a_{n-1}u_{n-1}$. But $1-a_n$ is a unit, since $a_n\in J(A)$, hence $u_n\in\langle u_1,\dots,u_{n-1}\rangle$. Iterating, we see that all $u_i$ have been zero.

Alternative proof of Nak1: Let $S=1+\mathfrak{a}$. Then $S^{-1}\mathfrak{a}\subseteq J(S^{-1}A)$. If $M=\mathfrak{a}M$, then $S^{-1}M=S^{-1}(\mathfrak{a}M)=(S^{-1}\mathfrak{a})(S^{-1}M)$, thus Nak2 implies $S^{-1}M=0$. As $M$ is finitely generated, there is an $x\in S$ such that $xM=0$.

Now for my question:

Can CH be deduced from Nakayama's Lemma, avoiding linear algebra, in particular the theory of determinants?

By the way, the arguments are taken from Atiyah-Macdonald, I did not find them myself.

• If there exists a non-constant polynomial in $\phi$ which is surjective and $M$ is finitely generated, then yes. $M$ is naturally an $A[x]$ module with $x$ acting as $\phi$. Then $p(\phi)M=M$ for some $p$ so there exists $q[x] \in A[x]$ with $q(\phi)M=0$. For certain cases (i.e. when $A=\Bbb R$), this is enough as $kI+\phi$ is always invertible for sufficiently large $k$. – PVAL-inactive Jul 7 '15 at 22:31
• Of course you can't avoid linear algebras, as modules themselves are part of linear algebra. the question is, as it stands, ill-posed.. – Mister Benjamin Dover Jul 7 '15 at 22:31
• @BenDover then, why are some books called "homological algebra", others "commutative algebra" and some are called "linear algebra"? I don't think linear algebra can be defined as the theory of modules. – Jakob Werner Jul 7 '15 at 22:35
• @JakobWerner it is very simple: because they deal with different aspects of modules (e.g. commutative algebra deals a lot with "local" properties of modules, and solely with modules over commutative rings; classical homological algebra with derived functors of functors on categories of modules over a not necessairyl commutative ring). Ofc these subjects are all related, and linear algebra is the most basic subject of these. I never said that linear algebra "is" the theory of modules, but that the notion of a "module" is a notion of linear algebra. many people agree with this, e.g. bourbaki. – Mister Benjamin Dover Jul 7 '15 at 23:03
• The point of CH is not that $\phi$ satisfies is a polynomial equation but that it satisfies is a polynomial equation of degree $n$. See math.stackexchange.com/q/1310609/589. – lhf Jul 8 '15 at 13:05

## 1 Answer

Nakayama Lemma. Let $N$ be a finitely generated $R$-module, and $J\subseteq R$. Suppose that $J$ is closed under addition and multiplication and $JN=N$. Then there is $a\in J$ such that $(1+a)N=0$. (Here by $JN$ we denote the subset of linear combinations of $N$ with coefficients in $J$.)

Cayley-Hamilton Theorem. Let $A$ be a commutative ring, $I$ an ideal of $A$, $M$ a finitely generated $A$-module, $\phi$ an $A$-module endomorphism of $M$ such that $\phi(M)\subseteq IM$. Then there are $n\ge 1$ and $a_i\in I^i$ such that $\phi^n+a_1\phi^{n-1}+\dots+a_n=0$.

Nakayama Lemma implies Cayley-Hamilton Theorem:

$M$ is an $A[X]$-module via $Xm=\varphi(m)$. Moreover, $M$ is also a finitely generated $A[X]$-module. By hypothesis $XM\subseteq IM$. Now consider the ring $A[X,X^{-1}]$, the localization of $A[X]$ with respect to the multiplicative set $S$ generated by $X$, and the finitely generated $A[X,X^{-1}]$-module $S^{-1}M$ (which we denote by $M[X^{-1}]$). The set $$J=\{a_1X^{-1}+\cdots+a_{r}X^{-r}:a_i\in I^i, r\ge1\}$$ is closed under addition and multiplication, and moreover $JM[X^{-1}]=M[X^{-1}]$: if $m\in M$ and since $XM\subseteq IM$ we have $Xm\in IM$. Then $Xm=a_1m_1+\cdots+a_nm_n$ with $a_j\in I$, and therefore $m=(a_1X^{-1})m_1+\cdots+(a_nX^{-1})m_n\in JM[X^{-1}]$.

Now by Nakayama Lemma (for $R=A[X,X^{-1}]$ and $N=M[X^{-1}]$) there is $p\ge 1$ such that $(1+a_1X^{-1}+\cdots+a_{p}X^{-p})M[X^{-1}]=0$. In particular, $(1+a_1X^{-1}+\cdots+a_{p}X^{-p})M=0$, that is, $\dfrac{(X^p+a_1X^{p-1}+\cdots+a_p)m}{X^p}=0$ for all $m\in M$. Since $M$ is finitely generated there is $s\ge0$ such that $X^s(X^p+a_1X^{p-1}+\cdots+a_p)M=0$. Now set $n=s+p$ and conclude that $\varphi^{n}+a_1\varphi^{n-1}+\cdots+a_n=0$ with $a_i\in I^i$.

• Basically right, but two corrections. The equation $(X^n+a_1 X^{n-1} + \cdots + a_n) M = 0$ is true in $M[X^{-1}]$. Since $M$ need not inject into $M[X^{-1}]$, this doesn't prove the claim in $M$. However, you do get to deduce $(X^n+a_1 X^{n-1} + \cdots + a_n) M \subset \mathrm{Ker}(M \to M[X^{-1}])$ and thus (since $M$ is finitely generated) $X^s (X^n+a_1 X^{n-1} + \cdots + a_n) M=0$ for some $s$. – David E Speyer Jul 8 '15 at 13:52
• Also, the name of your ring switched from $R$ to $A$ between the theorem statement and the proof. – David E Speyer Jul 8 '15 at 13:52
• @DavidSpeyer Thanks for the comments. I've just edited the initial proof, but it seems this wasn't the best choice. (Actually I wanted to obtain $a_i\in I^i$ which is one of the common conclusions of C-H.) – user26857 Jul 8 '15 at 13:59
• Unless I am mistaken, your version of Nakayama is not true. See my answer at math.stackexchange.com/questions/2102320/…. Beware that if $J$ is not an ideal, $JN=N$ does not imply that if $S$ generates $N$ then every element of $N$ is a $J$-linear combination of elements of $S$, which breaks many of the usual proofs of Nakayama. – Eric Wofsey Jan 17 '17 at 23:58
• Anyway, it seems I've aimed to a sharper version of Cayley-Hamilton. If one restricts to the version stated by the OP and use $IA[X,A^{-1}]$ instead of $J$, then we are done. (I don't edit the answer for now, but I'll do it soon.) – user26857 Jan 18 '17 at 9:42